how can I prove with induction this: $\binom{2n}{0}\le \binom{2n}{1}\le\binom{2n}{2}\le \cdots\le \binom{2n}{n}$?

Question

how can I prove: $\binom{2n+2}{0}\le \binom{2n+2}{1}\le \cdots\le \binom{2n+2}{n+1}$ relying this $\binom{2n}{0}\le \binom{2n}{1}\le\binom{2n}{2}\le \cdots\le \binom{2n}{n}$?

Answer 1

Removing one term in one end of your inequality chain gives us these two chains of inequalities: $$ \binom{2n}{0}\le \binom{2n}{1}\le\binom{2n}{2}\le \cdots\le \binom{2n}{n-1}\\ \binom{2n}{1}\le\binom{2n}{2}\le \binom{2n}3\le\cdots\le \binom{2n}{n} $$ Adding these term-wise gives us $$ \binom{2n + 1}{1}\le \binom{2n + 1}{2}\le\binom{2n + 1}{3}\le \cdots\le \binom{2n+1}{n} $$ It is easy to see that we can tack on $\binom{2n+1}0\le{}$ on the left side and ${}\le\binom{2n+1}{n+1}$ on the right side (the first one because $1\le 2n+1$, and the second one because $\binom{2n+1}n = \binom{2n+1}{n+1}$). Now do the same addition trick again, and you're basically done.

Answer 2

Hint $$\binom{2n+2}{k}=\binom{2n+1}{k}+\binom{2n+1}{k-1}=\binom{2n}{k}+\binom{2n}{k-1}+\binom{2n}{k-1}+\binom{2n}{k-2}\\ =\binom{2n}{k}+2\binom{2n}{k-1}+\binom{2n}{k-2}$$

Pay extra attention to $k=0,1$ amd $n=n+1, n+2$.

Answer 3

Here is a direct proof, no need for induction :

Let $ n $ be a positive integer, since $ \left(\forall k\in\left[\!\left[0,n-1\right]\!\right]\right),\ 2n-2k-1\geq 0 $, we have the following : $$ \left(\forall k\in\left[\!\left[0,n-1\right]\!\right]\right),\ \binom{2n}{k+1}-\binom{2n}{k}=\left(\frac{2n-k}{k+1}-1\right)\binom{2n}{k}=\frac{2n-2k-1}{k+1}\binom{2n}{k}\geq 0 $$

Thus, $$ \left(\forall k\in\left[\!\left[0,n-1\right]\!\right]\right),\ \binom{2n}{k}\leq\binom{2n}{k+1} $$

Answer 4

Let prove that : $r<k\leq n$

$$\binom{2n}{r}\leq\binom{2n}{k}$$

We must show :

$$\frac{(2n)!}{r!(2n-k)!}\leq \frac{(2n)!}{k!(2n-k)!}$$

Or

$$k!(2n-k)!\leq r!(2n-r)!$$

For $ n=0$ true For $n=1$ , $k=1,r=0$

$1!(1)!\leq 1!2!$

For n+1 then:

$$k!(2n+1-k)!=k!(2n-k)!(2n+1-k)\leq r!(2n-r)!(2n+1-r)$$

Proved!

Its celery $2n+1-k\leq 2n+1-r$